Question

When a metal was exposed to one photon of light at a frequency of 4.26× 1015 s–1, one electron was emitted with a kinetic energy of 5.00× 10–19 j. calculate the work function of this metal?

Answer 1

The electron's energy when it leaves the metal is equal to the energy of the photon that hit it minus the energy required to break free of the metal (the work function) The photon's energy was


`E_( \gamma)=hf=6.626 * 10^-^3^4Js * 4.26 * 10^1^5 s^-^1=2.823* 10^-^1^8J`

The work function is the difference between this and the electron's kinetic energy:


`\Phi =2.823* 10^-^1^8J-5.00* 10^-^1^9J=2.323* 10^-^1^8J`